Problem: Let $g(x)=\dfrac{x-5}{\sqrt{x-4}-1}$ when $x\neq 5$. $g$ is continuous for all $x>4$. Find $g(5)$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $8$ (Choice C) C $5$ (Choice D) D $10$
Explanation: $\dfrac{x-5}{\sqrt{x-4}-1}$ is continuous for all $x>4$ other than $x=5$, which means $g$ is continuous for all $x>4$ other than $x=5$. In order for $g$ to also be continuous at $x=5$, the following equality must hold: $\lim_{x\to 5}g(x)=g(5)$ We will obtain the above equality by letting $g(5)=\lim_{x\to 5}g(x)$. So let's find $\lim_{x\to 5}g(x)$, come on! $\begin{aligned} &\phantom{=}\lim_{x\to 5}g(x) \\\\ &=\lim_{x\to 5}\dfrac{x-5}{\sqrt{x-4}-1} \gray{\text{This is the rule for }x\neq 5} \\\\ &=\lim_{x\to 5}\dfrac{x-5}{\sqrt{x-4}-1}\cdot\dfrac{\sqrt{x-4}+1}{\sqrt{x-4}+1} \gray{\text{Rationalize}} \\\\ &=\lim_{x\to 5}\dfrac{(x-5)(\sqrt{x-4}+1)}{x-4-1^2} \gray{\text{Simplify}} \\\\ &=\lim_{x\to 5}\dfrac{\cancel{(x-5)}(\sqrt{x-4}+1)}{\cancel{x-5}} \gray{\text{Cancel common factors}} \\\\ &=\lim_{x\to 5}(\sqrt{x-4}+1) \\\\ &\text{(This is allowed because }x\neq 5) \\\\ &=\sqrt{5-4}+1 \gray{\text{Direct substitution}} \\\\ &=2 \end{aligned}$ We obtained that if we set $g(5)=2$, then $\lim_{x\to 5}g(x)=g(5)$, which makes $g$ continuous at $x=5$. Since we already saw that $g$ is continuous for any other $x>4$, we can determine that it's continuous for all $x>4$. In conclusion, $k=2$.